POJ-3292 Semi-prime H-numbers

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8564   Accepted: 3739

Description

This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.

An H-number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,... are the H-numbers. For this problem we pretend that these are the only numbers. The H-numbers are closed under multiplication.

As with regular integers, we partition the H-numbers into units, H-primes, and H-composites. 1 is the only unit. An H-number h is H-prime if it is not the unit, and is the product of two H-numbers in only one way: 1 × h. The rest of the numbers are H-composite.

For examples, the first few H-composites are: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 = 85.

Your task is to count the number of H-semi-primes. An H-semi-prime is an H-number which is the product of exactly two H-primes. The two H-primes may be equal or different. In the example above, all five numbers are H-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it's the product of three H-primes.

Input

Each line of input contains an H-number ≤ 1,000,001. The last line of input contains 0 and this line should not be processed.

Output

For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated by one space in the format shown in the sample.

Sample Input

21 
85
789
0

Sample Output

21 0
85 5
789 62

Source

Waterloo Local Contest, 2006.9.30

题目大意:定义了一类数的集合(膜4余1),并定义了相应的素数和合数,问1到h中半素数有多少。


分析:仿照一般的筛法求出1到1000001之间的H素数,然后离线暴力求出所有的半素数,O(1)处理询问。


#include <cstdio>
#include <cmath>
#include <iostream>
#define MAXN 1000001 
using namespace std;
int x,num,prim[MAXN],sum[MAXN];
bool inprim[MAXN+1];
int Sqrt(int x)
{
	return int(sqrt(double(x)));
}
int main()
{
	for(int i = 5;i <= MAXN;i += 4)
	 if(!inprim[i])
	 {
		for(int j = 5;j*i <= MAXN;j += 4) 
		 inprim[i*j] = true; 
		prim[++num] = i;
	 }
	for(int i = 5;i <= MAXN;i += 4)
	{
		for(int j = 1;j <= num && prim[j] <= Sqrt(i)+1;j++)
		 if(i % prim[j] == 0) 
		 {
		 	if(!inprim[i / prim[j]]) sum[i] = 1;
		 	break;
		 }
	}
	for(int i = 1;i <= MAXN;i++)
	 sum[i] += sum[i-1];
	cin.sync_with_stdio(false);
	while(cin>>x && x)
	{
		cout<<x<<" "<<sum[x]<<endl;
	}
} 


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